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# Find the sum of all two digit numbers which when divided by $4$ yields $1$ as remainder.

$\begin{array}{1 1}1196 \\ 1210 \\ 1312 \\ 1218 \end{array}$

Toolbox:
• $n^{th}$ term of an A.P.=$a+(n-1)d$
• Sum of $n$ terms of an A.P=$\large\frac{n}{2}$$(t_n+a) Two digit numbers that yields remainder 1 when divided by 4 forms sequence 13,17,21,........97 This sequence is an A.P with first term a=13 common difference d=4 and t_n=97 \Rightarrow\:a+(n-1)d=97 \Rightarrow\:13+(n-1)4=97 \Rightarrow\:n-1=\large\frac{84}{4}$$=21$
or $n=22$
$\therefore$ Sum of the required numbers =$13+17+21+.......97$
$=\large\frac{n}{2}$$(t_n+a)=11(97+13)=11\times 110=1210$